I need to create a predicate that will look for the following string:
"fred\n5" where \n is a newline.
At least, this is string that is returned when reading the metadata back
You can do it with Regular Expression
let string = """
fred
5
"""
let predicate = NSPredicate(format: "self MATCHES %#", "fred\\n5")
predicate.evaluate(with: string) // true
It's also possible to use the pattern fred(\\n|\\r)5, it considers both linefeed and return.
Alternatively remove the newline character (actually any whitespace and newline characters)
let trimmedString = string.replacingOccurrences(of: "\\s", with: "", options: .regularExpression)
I am saving users input to db as a string and I would like to remove all spaces at each lines.
Input from user:
Hi!
My name is:
Bob
I am from the USA.
I want to remove spaces between "Bob", so the result will be:
Hi!
My name is:
Bob
I am from the USA.
I am trying to do it with the following code
let regex = try! NSRegularExpression(pattern: "\n[\\s]+", options: .caseInsensitive)
a = regex.stringByReplacingMatches(in: a, options: [], range: NSRange(0..<a.utf16.count), withTemplate: "\n")
but this code replace multiple new lines "\n", I don't want to do it.
After I run the above code: "1\n\n\n 2" -> "1\n2". The result I need: "1\n\n\n2" (only spaces are removed, not new lines).
No need for regex, split the string on the new line character into an array and then trim all lines and join them together again
let trimmed = string.components(separatedBy: .newlines)
.map { $0.trimmingCharacters(in: .whitespaces) }
.joined(separator: "\n")
or you can use reduce
let trimmed = string.components(separatedBy: .newlines)
.reduce(into: "") { $0 += "\($1.trimmingCharacters(in: .whitespaces))\n"}
You can use
let regex = try! NSRegularExpression(pattern: "(?m)^\\h+", options: .caseInsensitive)
Actually, as there are no case chars in the pattern, you may remove .caseInsensitive and use:
let regex = try! NSRegularExpression(pattern: "(?m)^\\h+", options: [])
See the regex demo. The pattern means:
(?m) - turn on multiline mode
^ - due to (?m), it matches any line start position
\h+ - one or more horizontal whitespaces.
Swift code example:
let txt = "Hi!\n\nMy name is:\n Bob\n\nI am from the USA."
let regex = "(?m)^\\h+"
print( txt.replacingOccurrences(of: regex, with: "", options: [.regularExpression]) )
Output:
Hi!
My name is:
Bob
I am from the USA.
I'm bit confused by NSRegularExpression in swift, can any one help me?
task:1 given ("name","john","name of john")
then I should get ["name","john","name of john"]. Here I should avoid the brackets.
task:2 given ("name"," john","name of john")
then I should get ["name","john","name of john"]. Here I should avoid the brackets and extra spaces and finally get array of strings.
task:3 given key = value // comment
then I should get ["key","value","comment"]. Here I should get only strings in the line by avoiding = and //
I have tried below code for task 1 but not passed.
let string = "(name,john,string for user name)"
let pattern = "(?:\\w.*)"
do {
let regex = try NSRegularExpression(pattern: pattern, options: .caseInsensitive)
let matches = regex.matches(in: string, options: [], range: NSRange(location: 0, length: string.utf16.count))
for match in matches {
if let range = Range(match.range, in: string) {
let name = string[range]
print(name)
}
}
} catch {
print("Regex was bad!")
}
Thanks in advance.
RegEx in Swift
These posts might help you to explore regular expressions in swift:
Does a string match a pattern?
Swift extract regex matches
How can I use String slicing subscripts in Swift 4?
How to use regex with Swift?
Swift 3 - How do I extract captured groups in regular expressions?
How to group search regular expressions using swift?
Task 1 & 2
This expression might help you to match your desired outputs for both Task 1 and 2:
"(\s+)?([a-z\s]+?)(\s+)?"
Based on Rob's advice, you could much reduce the boundaries, such as the char list [a-z\s]. For example, here, we can also use:
"(\s+)?(.*?)(\s+)?"
or
"(\s+)?(.+?)(\s+)?"
to simply pass everything in between two " and/or space.
RegEx
If this wasn't your desired expression, you can modify/change your expressions in regex101.com.
RegEx Circuit
You can also visualize your expressions in jex.im:
JavaScript Demo
const regex = /"(\s+)?([a-z\s]+?)(\s+)?"/gm;
const str = `"name","john","name of john"
"name"," john","name of john"
" name "," john","name of john "
" name "," john"," name of john "`;
const subst = `\n$2`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Task 3
This expression might help you to design an expression for the third task:
(.*?)([a-z\s]+)(.*?)
const regex = /(.*?)([a-z\s]+)(.*?)/gm;
const str = `key = value // comment
key = value with some text // comment`;
const subst = `$2,`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Separate the string by non alpha numeric characters except white spaces. Then trim the elements with white spaces.
extension String {
func words() -> [String] {
return self.components(separatedBy: CharacterSet.alphanumerics.inverted.subtracting(.whitespaces))
.filter({ !$0.isEmpty })
.map({ $0.trimmingCharacters(in: .whitespaces) })
}
}
let string1 = "(name,john,string for user name)"
let string2 = "(name, john,name of john)"
let string3 = "key = value // comment"
print(string1.words())//["name", "john", "string for user name"]
print(string2.words())//["name", "john", "name of john"]
print(string3.words())//["key", "value", "comment"]
Here I have done with after understanding all of above comments.
let text = """
Capturing and non-capturing groups are somewhat advanced topics. You’ll encounter examples of capturing and non-capturing groups later on in the tutorial
"""
extension String {
func rex (_ expr : String)->[String] {
return try! NSRegularExpression(pattern: expr, options: [.caseInsensitive])
.matches(in: self, options: [], range: NSRange(location: 0, length: self.count))
.map {
String(self[Range($0.range, in: self)!])
}
}
}
let r = text.rex("(?:\\w+-\\w+)") // pass any rex
A single pattern, works for test:1...3, in Swift.
let string =
//"(name,john,string for user name)" //test:1
//#"("name"," john","name of john")"# //test:2
"key = value // comment" //test:3
let pattern = #"(?:\w+)(?:\s+\w+)*"# //Swift 5+ only
//let pattern = "(?:\\w+)(?:\\s+\\w+)*"
do {
let regex = try NSRegularExpression(pattern: pattern)
let matches = regex.matches(in: string, range: NSRange(0..<string.utf16.count))
let matchingWords = matches.map {
String(string[Range($0.range, in: string)!])
}
print(matchingWords) //(test:3)->["key", "value", "comment"]
} catch {
print("Regex was bad!")
}
Let’s consider:
let string = "(name,José,name is José)"
I’d suggest a regex that looks for strings where:
It’s the substring either after the ( at the start of the full string or after a comma, i.e., look behind assertion of (?<=^\(|,);
It’s the substring that does not contain , within it, i.e., [^,]+?;
It’s the substring that is terminated by either a comma or ) at the end of the full string, i.e., look ahead assertion of (?=,|\)$), and
If you want to have it skip white space before and after the substrings, throw in the \s*+, too.
Thus:
let pattern = #"(?<=^\(|,)\s*+([^,]+?)\s*+(?=,|\)$)"#
let regex = try! NSRegularExpression(pattern: pattern)
regex.enumerateMatches(in: string, range: NSRange(string.startIndex..., in: string)) { match, _, _ in
if let nsRange = match?.range(at: 1), let range = Range(nsRange, in: string) {
let substring = String(string[range])
// do something with `substring` here
}
}
Note, I’m using the Swift 5 extended string delimiters (starting with #" and ending with "#) so that I don’t have to escape my backslashes within the string. If you’re using Swift 4 or earlier, you’ll want to escape those back slashes:
let pattern = "(?<=^\\(|,)\\s*+([^,]+?)\\s*+(?=,|\\)$)"
I am trying to determine whether an input string contains "n't" or "not".
For example, if the input were:
let part = "Hi, I can't be found!"
I want to find the presence of the negation.
I have tried input.contains, .range, and NSRegularExpression. All of these succeed in finding "not", but fail to find "n't". I have tried escaping the character as well.
'//REGEX:
let negationPattern = "(?:n't|[Nn]ot)"
do {
let regex = try NSRegularExpression(pattern: negationPattern)
let results = regex.matches(in: text,range: NSRange(part.startIndex..., in: part))
print("results are \(results)")
negation = (results.count > 0)
} catch let error {
print("invalid regex: \(error.localizedDescription)")
}
//.CONTAINS
if part.contains("not") || part.contains("n't"){
print("negation present in part")
negation = true
}
//.RANGE (showing .regex option; also tried without)
if part.lowercased().range(of:"not", options: .regularExpression) != nil || part.lowercased().range(of:"n't", options: .regularExpression) != nil {
print("negation present in part")
negation = true
}
Here is a picture:
This is a bit tricky, and the screenshot is actually what gives it away: your regex pattern has a plain single quote in it, but the input text has a "smart" or "curly" apostrophe in it. The difference is subtle:
Regular: '
Smart: ’
Lots of text fields will automatically replace regular single quotes with "smart" apostrophes when they think it's appropriate. Your regex, however, only matches the plain single quote, as evidenced by this tiny test:
func isNegation(input text: String) -> Bool {
let negationPattern = "(?:n't|[Nn]ot)"
let regex = try! NSRegularExpression(pattern: negationPattern)
let matches = regex.matches(in: text,range: NSRange(text.startIndex..., in: text))
return matches.count > 0
}
for input in ["not", "n't", "n’t"] {
print("\"\(input)\" is negation: \(isNegation(input: input) ? "YES" : "NO")")
}
This prints:
"not" is negation: YES
"n't" is negation: YES
"n’t" is negation: NO
If you want to continue using a regex for this problem, you'll need to modify it to match this kind of punctuation character, and avoid assuming all your input text includes "plain" single quotes.
Using swift 4, I need to parse a string to get substrings that will always be different. For example:
let str = "[33376:7824459] Device Sarah's Phone (Hardware: D21AP, ECID: 8481036056622, UDID: 76e6bc436fdcfd6c4e39c11ed2fe9236bb4ec, Serial: F2LVP5JCLY)"
let strRange = str.range(of: "(?<=Device )(?= (Hardware)", options: .regularExpression)
print(strRange!)
I would think this would output "Sarah's Phone"
I'm not getting any errors on this code, but it's also not working. What am I doing wrong?
Several problems:
You have a lookahead and lookbehind here, but nothing that would actually match any characters, so it'll never match anything except an empty string.
You didn't properly escape the parenthesis in your lookahead.
You should use if let or guard let, rather than !, to unwrap the optional. Otherwise, you'll get a crash when you encounter an input string that doesn't match the pattern.
I'm not sure why you'd expect print(strRange) to output text. strRange is a range, not a string or a substring.
This sample will fix your problems:
import Foundation
let str = "[33376:7824459] Device Sarah's Phone (Hardware: D21AP, ECID: 8481036056622, UDID: 76e6bc436fdcfd6c4e39c11ed2fe9236bb4ec, Serial: F2LVP5JCLY)"
if let strRange = str.range(of: "(?<=Device ).*(?= \\(Hardware)", options: .regularExpression) {
print(str[strRange])
} else {
print("Not Found")
}